递归编写一个十进制转换为二进制的函数
def Dec2Bin(dec):result = ''
if dec:
result = Dec2Bin(dec//2)
return result + str(dec%2)
else:
return result
print(Dec2Bin(62))
下面这两句我不太理解：最好能分解到每一步？谢谢！！
result = Dec2Bin(dec//2)
return result + str(dec%2) def Dec2Bin(dec):
result = ''
if dec: #如果dec不等于0
result = Dec2Bin(dec//2)
return result + str(dec%2)
'''
result + str(dec%2)=Dec2Bin(62//2) + str(62%2)
=Dec2Bin(31) + str(62%2)
=Dec2Bin(31//2) +str(31%2) + str(62%2)
=Dec2Bin(15) +str(31%2) + str(62%2)
=Dec2Bin(15//2) + str(15%2) +str(31%2) + str(62%2)
=Dec2Bin(7) + str(15%2) +str(31%2) + str(62%2)
=Dec2Bin(7//2) + str(7%2) + str(15%2) +str(31%2) + str(62%2)
=Dec2Bin(3) + str(7%2) + str(15%2) +str(31%2) + str(62%2)
=Dec2Bin(3//2) + str(3%2) + str(7%2) + str(15%2) +str(31%2) + str(62%2)
=Dec2Bin(1) + str(3%2) + str(7%2) + str(15%2) +str(31%2) + str(62%2)
=Dec2Bin(1//2)+ str(1%2) + str(3%2) + str(7%2) + str(15%2) +str(31%2) + str(62%2)
=Dec2Bin(0)+ str(1%2) + str(3%2) + str(7%2) + str(15%2) +str(31%2) + str(62%2)
='' +'1'+'1'+'1'+'1'+'1'+'0'
='111110'
'''
else:
return result
print(Dec2Bin(62))
看了其他帖子，，，，理解了 先对函数变形，效果一样，但是更容易理解
def Dec2Bin(dec):
if dec:
return Dec2Bin(dec // 2) + str(dec % 2)
else:
return ''
下面是递归的全过程：
Dec2Bin(62) = Dec2Bin(62 // 2) + str(62 % 2) = Dec2Bin(31) + '0'
Dec2Bin(31) = Dec2Bin(31 // 2) + str(31 % 2) = Dec2Bin(15) + '1'
Dec2Bin(15) = Dec2Bin(15 // 2) + str(15 % 2) = Dec2Bin( 7) + '1'
Dec2Bin( 7) = Dec2Bin( 7 // 2) + str( 7 % 2) = Dec2Bin( 3) + '1'
Dec2Bin( 3) = Dec2Bin( 3 // 2) + str( 3 % 2) = Dec2Bin( 1) + '1'
Dec2Bin( 1) = Dec2Bin( 1 // 2) + str( 1 % 2) = Dec2Bin( 0) + '1'
Dec2Bin( 0) = ''
只要从最下面把 Dec2Bin(0) = '' 带入 Dec2Bin(1)，得到 Dec2Bin(1) = '1' , 再把这个结果带入 Dec2Bin(3) 得到 Dec2Bin(3) = '11' 。。。。。。最终带入的结果，Dec2Bin(62) = '111110'
Dec2Bin(62) = '' + '1' + '1' + '1' + '1' + '1' + '0' = '111110'
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